Diophantine approximation with improvement of the simultaneous control of the error and of the denominator

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Diophantine approximation with improvement of the simultaneous control of the error and of the denominator

Abdelmadjid Boudaoud

To cite this version:

Abdelmadjid Boudaoud. Diophantine approximation with improvement of the simultaneous control of the error and of the denominator. 2016. �hal-01312603�

Diophantine approximation with

improvement of the simultaneous control of the error and of the denominator

Abdelmadjid BOUDAOUD Department of Mathematics,

Faculty of Mathematics and Computer Sciences, University of M’sila, Algeria

Laboratory of Pure and Applied Mathematics (L.M.P.A.) May 7, 2016

Abstract

In this work we proof the following theorem which is, in addition to some other lemmas, our main result:

theorem. Let X =f(x1,t1), (x2,t2), ..., (xn,tn)g be a …nite part of R R ⁺, then there exist a …nite part R of R ⁺ such that for all

" >0there existsr 2Rsuch that if0< " rthen there exist rational numbers pi

q i=1;2;:::;n

such that:

8<

:

x_i p_i

q "t_i

"q t_i

,i= 1;2; :::; n. (*) It is clear that the condition "q ti fori= 1;2; :::; n is equivalent to

"q t = M in

i=1;2;:::;n (t_i). Also, we have (*) for all " verifying 0 < "

"₀= minR.

The previous theorem is the classical equivalent of the following one which is formulated in the context of the nonstandard analysis ([2],[5],[6],[8]).

theorem. For every positive in…nitesimal real ", there exists an un-

( x = px

q +"

"q = 0

.

For this reason, to prove the nonstandard version of the main result and to get its classical version we place ourselves in the context of the nonstandard analysis.

1991 Mathematics Subject Classi…cation. 11J13, 03H05, 26E35.

Key words and phrases. Diophantine approximation, Farey series, Non- standard Analysis.

1 Introduction, Notations and Rappel

We dispose in the domain of Diophantine approximation of many results (refer for example to [3], [7]). In the following, we give as an example, the two most used theorems:

Theorem (Dirichlet) 1.1. [7]. Suppose that x₁; x₂; ::: ;x_n are n real numbers and that T >1is an integer. Then there exist integers q,p1,p2,:::,pn

with

8<

:

x_i p_i q

1

T q (i= 1;2; :::; n) 1 q < Tⁿ

. (1.1)

Theorem (Kronecker) 1.2. [7]. For any reals ₁; ₂; ::: ; _nand anyt >0, the system of inequalities

8>

><

>>

:

jq ₁ p_{1 1}j< t jq ₂ p_{2 2}j< t

::::::::::::::::::

jq _n p_{n n}j< t

(1.2)

is solvable in integers q, p₁; p₂; ::: ;p_n if and only if ₁; ₂; ::: ; _n are not rationally dependent. Note that ₁; ₂; ::: ; _n are said rationally dependent if there exist integers r, r₁; r₂; ::: ;r_n not all zero such that

r_{1 1} +r_{2 2}+:::+r_{n n} =r.

When we take ₁ = ₂ =::: = _n= 0, this theorem is used to approximate the reals _i by using rationals p_i

q to errors smaller than t q.

In general, in these results we observe that the simultaneous control be- tween the error and the common denominatorq should be clari…ed and spec- i…ed. This, because the approximation to a given error (which is generally small) requires a denominator that is generally too big. Conversely, the ap- proximation with a small denominator might give an error that is not really small. This question has motivated us to give the following theorem which is, in addition to some other lemmas, our main result of this work.

Theorem 1.3. Let X = f(x₁,t₁), (x₂, t₂), ..., (x_n, t_n)g be a …nite part of R R ⁺, then there exist a …nite part R of R ⁺ such that for all " > 0 there exists r 2 R such that if 0< " r then there exist rational numbers

p_i

q i=1;2;:::;n

such that:

8<

:

x_i p_i

q "t_i

"q ti

, i= 1;2; :::; n. (1.3) We note that in (1.3) the condition "q t_i for i= 1;2; :::; n is equivalent to

"q t = M in

i=1;2;:::;n (t_i). Also, under the assumption of theorem 1.3, for all "

verifying0< " "₀ = minR we obtain (1.3).

The theorem 1.3 is the classical equivalent of the following theorem (the- orem 1.4.) formulated in the context of the nonstandard analysis.

Theorem 1.4. For every positive in…nitesimal real", there exists an integer Q depending only of ", such that8^stx2R 9 P_x 2Z:

8<

:

x = P_x

Q +"

"Q = 0

. (1.4)

In the following we make a comparison between our result (theorem 1.3) and the existing results such as Dirichlet’s theorem and Kronecker’s theorem.

Our main result is used to approximate at a reduced common denominator q since "q t (i.e. q t

") and at a di¤erent errors since x_i pi

q "t_i for i= 1;2; :::; n. In addition, if we taket1 =t2 =:::=tn=t >0and "0 =M in R then for every 0< " " there exist integers q,p ,p ,:::,p such that

M ax

i2f1;2;:::;ng

x_i p_i

q "tand q t

" (1.5)

i.e., a denominatorq t

" enough for an error not exceeding "t.

Look when we use, under the same hypotheses, the Dirichlet’s theorem.

It may happen that when we take 1

T > "t, the common denominator q 1 is small enough so that the maximum error is strictly greater than "t i.e.

"t < M ax

i2f1;2;:::;ng x_i p_i q

1 T q

1

T. In contrast, when we take T satisfying 1

T "t then we are sure that the maximum error is smaller than or equal to"ti.e. M ax

i2f1;2;:::;ng x_i p_i q

1 T q

1

T "t. But in this case it may happen that the common denominator q, since that 1 q < Tⁿ, is very close to

Tⁿ 1

("t)ⁿ ( q = Tⁿ 1 1

("t)ⁿ 1; for instance). Consequently, to be sure of the realization of the approximation asked, it is necessary to choose

1

T "tand q can be too big in this case as we have seen.

On his part the Kronecker’s theorem is purely existential and don’t say anything on the common denominator.

From the above we can see that the theorem 1.3 ensure the ability to control the size of q and of the maximum error; especially when " (resp. n) become small (resp. large). For its proof we place ourselves in the framework of the nonstandard analysis and we proceed as follows :

(1) We …rst show theorem 1.4 (In the sequel noted theorem 2.1.) by using some lemmas.

(2) We translate theorem 1.4 by using the Nelson’s algorithm.

1.1 Notations

i) For a number x (integer or non) we have the following usages:

1) Abbreviation, st(x) indicates thatx is standard; 8^stx signi…es 8x[st(x) =)::].

2) x= +1 ( resp. x= 0) signi…es thatx is a positive unlimited (resp. x an in…nitesimal). x >

= 0 signi…es thatx is an in…nitesimal real strictly positive.

3) $(resp. ) signi…es a limited real (resp. an in…nitesimal real) on which one doesn’t say anything besides.

4) kxk is the di¤erence, taken positively, betweenx and the nearest integer.

5) E(x) (resp. fxg) is the integral part of x (resp. the fractional part ofx;

that is fxg=x E(x)).

6) Let " be an in…nitesimal real, one designates by " galaxie(x) the set fy : y=x+"$g and by " halo(x) the setfy : y=x+" g.

7) x⁰ signi…es, forx limited, the standard part of x.

ii)

8) IfE is a given set,E (resp. jEj) designates the external set formed, only, by the standard elements of E (resp. the cardinality of E).

9) One notes by (x1; x2; :::; xn)^T the vector column 0 BB B@

x₁ x₂ ... x_n

1 CC CA.

1.2 Rappel

1.2.1 Farey series([3])

The Farey seriesF^N of orderN is the ascending series of irreducible fractions between 0 and 1 whose denominators do not exceed N. Thus h

k belongs to F^N if

0 h k N,(h; k) = 1 the numbers 0and 1are included in the forms 0

1 and 1 1. If h

k < h⁰ k⁰ < h⁰⁰

k⁰⁰ are three successive elements of F^N (N >1), then one has the following proper- ties:

1⁰) kh⁰ hk⁰ = 1.

2⁰) h⁰

k⁰ = h+h⁰⁰ k+k⁰⁰. 3⁰) k+k⁰ > N and h

k < h+h⁰ k+k⁰ < h⁰

k⁰.

4⁰) IfN >1, two successive elements ofF^N don’t have the same denominator.

5⁰) Let h1

k₁, h2

k₂ be two successive elements ofF^N (N 1) with h1

k₁ < h2

k₂, and let the two following sequences:8

><

>:

U₀ = h₂

k₂, U₁ = h₂+h₁

k₂+k₁, ... ,Ui = h₂+ih₁ k₂+ik₁, ...

V₀ = h₁

k₁, V₁ = h₁+h₂

k₁+k₂, ..., V_j = h₁+jh₂ k₁+jk₂, ...

. (1.6)

We prove easily that the sequence(U_i)_i2N (resp. (V_j)_j2N) is decreasing (resp.

increasing); besides we have:

8>

<

>:

U_i U_i+1 = 1

(k₂+ik₁) (k₂+ (i+ 1)k₁), U_i h₁

k₁ = 1

k₁(k₂ +ik₁)

V_j+1 V_j = 1

(k₁+jk₂) (k₁+ (j + 1)k₂), h2

k₂ V_j = 1

k₂(k₁ +jk₂) . (1.7) 1.2.2 Approximation to the in…nitesimal sense of reals

Theorem 1.5.[1]. Let be a real number. Then for all positive in…nitesimal real " there exist a rational number p

q and a limited real l such that:

( = p_i

q +"l

"q = 0 . (1.8)

2 Simultaneous approximation to the in…ni- tesimal sense of standard reals

We prove in this section the following theorem whose translation by the al- gorithm of Nelson gives the theorem 1.3 .

Theorem 2.1. For every positive in…nitesimal real", there existsan integer Q depending only of ", such that 8^stx2R9 P_x2Z:

8<

:

x = P_x

Q +"

"Q = 0

. (2.1)

Let" be a positive in…nitesimal real. We need to the following lemmas

Lemma 2.2. Let ( ₁; ₂; :::; _N) a system of real numbers with N 1 limited. Then for all positive in…nitesimal real there are rational numbers

p_i

q i=1;2;:::;N

and limited reals (l_i)i=1;2;:::;N such that fori= 1;2; :::; N : (

i = p_i

q + l_i

q = 0

. (2.2)

Proof. Consider, for everyn2N , the formula:

B(n) =

"8 ( ₁; ₂; :::; _n)2Rⁿ with n 1and 8 >

=

09 P_i

Q i=1;2;:::;n

such that for everyi2 f1;2; :::; ng : 8<

:

x_i P_i

Q = $

Q= 0

"

.

By theorem 1.5, we have B(1). Suppose, for 1 n a standard integer, B(n) and prove B(n+ 1). Let ₁; ₂; :::; _n; _n+1 2 Rⁿ⁺¹ and let >

= 0, then by B(n) there are rational numbers p_i

q i=1;2;:::;n

such that 8>

>>

>>

><

>>

>>

>>

:

1 = p₁

q + $

2 = p₂

q + $

... = ...

n = p_n

q + $

(2.3)

where q = 0. Now, since q = 0, the application of theorem 1.5 implies q _n+1 = p_n+1

q_n+1 + ( q)$,( q)q_n+1 = 0. Hence

n+1 = p_n+1

qq_n+1 + $, qq_n+1 = 0. (2.4)

We deduct from (2.3) and (2.4) that:

8>

>>

>>

>>

>>

>>

><

>>

>>

>>

>>

>>

>>

:

1 = p₁q_n+1

qq_n+1 + $ = P₁

Q + $

2 = p₂q_n+1

qq_n+1 + $ = P₂

Q + $

... = ... = ...

n = p_nq_n+1

qqn+1

+ $ = P_n

Q + $

n+1 = p_n+1

qq_n+1 + $ = P_n+1

Q + $

where, from (2.4), Q= qq_n+1 = 0. ConsequentlyB(n+ 1). Therefore, by the external recurrence principle, we have 8^stn 1 B(n).

Lemma 2.3. LetE be a given set. For all integer! = +1, there is a …nite subset F E containing all standard elements of E (i.e. E F ) and whose cardinal is strictly inferior to ! (jFj< !).

Proof. Let! = +1. LetB(F; z)be the internal formula: "F E,jFj< !, z 2 F ". Let Z E be a standard …nite part. Then there exists a …nite part F E with jFj < ! such that every element z of Z belongs to F, i.e.

we have B(F; z). Indeed it su¢ ces to take F =Z. Therefore, the principle of idealization (I) asserts the existence of a …nite part F E with jFj< ! such that any standard element of L belongs toF.

Lemma 2.4. Let = +1be a real number such that p

" = 0. LetF_M be the Farey sequence of orderM =E p

" . If p₁ q₁, p₂

q₂ are two elements of F_M such that q1 ' +1, q2 ' +1 and p₁

q₁ , p₂

q₂ doesn’t contain any standard rational number (in this case p₁

q₁ = p₂

q₂). Then there exist a …nite sequence of irreducible rational numbers l_i

m_i i=1;2;:::;g

such that:

p₁ q₁ = l₁

m₁ < l₂

m₂ < ::: < l_g m_g = p₂

q₂ where l_i+1

m_i+1 l_i

m_i = " for i = 1;2; :::; g 1. Besides for i = 1;2; :::; g we

have "m_i = 0and m_i = +1.

Proof. Let us consider the case where p₂ q₂

p₁

q₁ is not of " form; otherwise the lemma is proved. Let t_i

i i=1;2;:::;r

be the elements of F^M such that p₁

q₁ = t₁

1

< t₂

2

< ::: < t_r

r

= p₂ q₂. Let i₀ 2 f1;2; :::; r 1g such that t_i0+1

i0+1

t_i0

i0

is not of " form, because if a such i₀ does not exist the lemma is proved. From the properties of F^M ( 1.2.1), _i0+1 and _i0 cannot be equal. Then there are two cases:

A) _i

0+1 > _i

0 : Let us take, in this case, g₀ = +1 an integer such that g₀

i0

= 0 ( the existence of g0 is assured by Robinson’s lemma). Let X =

E g₀

" _i0 and

H = t_i0

i0

; U_p; U_{p 1}; :::; U₀ wherep=E X _i0+1

i0

and U_i = t_i0+1+i:t_i0

i0+1+i: _i0 (i= 0;1; :::; p 1; p). Now we prove that : p is an unlimited integer, the product of the denominator of every element of H by" is an in…nitesimal and the distance between two successive elements of H is of the " form.

Indeed, we haveX =E g₀

" _i

0

= g₀

" _i

0

X where _X 2[0;1[.

X _i0+1

i0

= g₀

" _i

0 i0

X i0

i0+1 i0

= g₀ " _i

0 X " _i

0 i0+1

" _{i0 i0} .

Since " _{i0 X} = 0, " _{i0 i0+1} is a limited real number otherwise t_i0+1

i0+1

t_i0

i0

= 1

i0 i0+1

=" what contradicts the supposition. Theng₀ " _i

0 X " _i

0 i0+1

is a positive unlimited real. On the other hand " _i

0 i0 is limited; then

X _i

0+1 i0

is a positive unlimited real, therefore p is also. The greatest denominator in H is _i0+1+p _i

0 where p= g₀

" _{i0 i0}

X i0

i0+1 i0

with 2[0;1[.

" _i0+1+p _i0 = " _i0+1+ g₀

" _{i0 i0}

X i0

i0+1 i0

i0

= " _i

0+1+ g₀

i0

" _X " _i

0+1 " _i

0 = 0.

Hence the product of the denominator of every element of H by " is an in…nitesimal. It remains to prove that the distance between two elements of H is of the " form; Indeed: Let i2 f0;1; :::; p 1g, from (1:7) we have

U_i U_i+1 = 1

i0+1+i: _{i0 i0+1}+ (i+ 1): _i0 .

By hypothesis we have _i0+1 > _i0, then of properties of Farey’s series (1.2.1)) 2 _i

0+1> _i

0+1+ _i

0 > M, then _i0+1 > M 2 . Let d_i =" _i

0+1+i: _i

0 i0+1+ (i+ 1): _i

0 . Seen that _i0+1 ² > M

2

2

, d_i is unlimited, therefore U_i U_i+1 =" . To

…nish the proof, we have of (1:7):

U_p t_i0

i0

= 1

i0+1+p _{i0 i0}.

Let d_p = " _i0+1+p: _{i0 i0}, after the replacement by the value of p, we obtain

d_p =" _{i0+1 i0} +g₀ " _{X i0} " _{i0+1 i0} " _{i0 i0}. Since " _i

0 = 0, " _i

0 i0 is limited, then d_p is unlimited; hence U_p ti0

i0

=" . Thus, we end what we perceived.

B) _i0 > _i0+1: Let us take, in this case, g₁ = +1 an integer such that g₁

i0+1

= 0 (the existence of g₁ is assured by Robinson’s lemma). Let Xe =

E g₁

" _i0+1 and

He = V₀; V₁; :::; V_p0 1; V_p0; t_i0+1

i0+1

where p⁰ = E Xe _i0

i0+1

!

and V_j = t_i0 +j:t_i0+1

i0 +j: _i0+1 ( j = 0;1; :::; p⁰ 1; p⁰).

Since the symmetry of this case with the case A) we prove, as in the case of H, that p⁰ is an unlimited integer, the product of the denominator of every element ofHe by"is an in…nitesimal and the distance between two successive elements of He is of the " form.

Thus the elements of H (or of He ) form a subdivision of the inter- val t_i0

i0

; t_i0+1

i0+1

. For the other intervals t_i

i

; t_i+1

i+1 i2f1;2;:::;r 1g fi0g

which don’t have a length of " form we do the same construction as we did with t_i0

i0

; t_i0+1

i0+1

.

By regrouping rational numbers which subdivide intervals ti i

; ti+1 i+1

(i 2 f1;2; :::; r 1g) not having a length of the " form and the rationals which are borders of intervals having a length of the " form, we obtain the

…nite sequence l_i

m_i i=1;2;:::;g

. The irreducibility of the elements of the se- quence li

m_i i=1;2;:::;g

results from properties of Farey’s series.

Lemma 2.5. Let 2[0 ,1]be a real, if is not in the"-galaxie of a standard rational number then there exists two irreducible rational numbers h₁

k₁;h₂ k₂ of the interval [0;1] such that

2 h1

k₁;h2

k₂ ; k₁ = +1; k₂ = +1; "k₁ ="k₂ = 0and h2

k₂ h1

k₁ =" .

such thatp

" = 0and letF_M be the Farey sequence of orderM =E p" . Let p₁

q₁, p₂

q₂ be two successive elements of FM such that 2 p₁ q₁;p₂

q₂ . Two cases are distinguished:

A) Nor p₁

q₁ nor p₂

q₂ is a standard rational : In this case by applying the lemma 2.4, we obtain two irreducible rationals l_i0

m_i0 and l_i0+1

m_i0+1 such that 2 l_i0

m_i0, l_i0+1

m_i0+1 , m_i0 = +1, m_i0+1 = +1, "m_i0 = "m_i0+1 = 0, l_i0+1 m_i0+1 l_i0

m_i0 =" . Hence the lemma is proved by taking l_i0

m_i0 for h₁

k₁ and l_i0+1

m_i0+1 for h₂ k₂. B) p₁

q₁ or p₂

q₂ is standard (cannot be both at the same time standard). Let us suppose that p₁

q₁ is standard (the other case, seen the symmetry, can be treated by the same way.). Then p₁

q₁ ="w where w = +1. Let us put

L=E 2= p₁

q₁ then "L= 0and p₁ q₁ + 1

L < . Let l

m be the reduced form of p₁

q₁ +1

L, then"m= 0because m Lq1 andq1 is a standard. m > M because l

m is not an element of F^M. Therefore "m² is an unlimited because

"m² > "M² and "M² is an unlimited. This means thatm is of theE

0

p"

!

form where ⁰ is a positive unlimited real verifying p

" ⁰ = 0. Now if we consider F^m, then 2 p⁰₁

q⁰₁;p⁰₂

q₂⁰ where p⁰₁

q₁⁰ and p⁰₂

q₂⁰ are two successive non standard elements ofF^m. Thus the caseB)comes back itself to the caseA), therefore the proposition is also proved for this case.

Remark. We easily see that this proof is also a proof for the theorem 1.5.

Let be a positive unlimited real such that": '0, then Lemma 2.6. There exists a …nite set

S =fl₁; l₂; :::; l_ng [0;1] (2.5) containing all standard elements of [0;1] such that jl_i+1 l_ij " for i 2 f1;2; :::; n 1g.

Proof. Let B(S; z) be the internal formula: "S [0;1] is …nite, z 2 S &

8(x₁ , x₂) 2 S S (jx₁ x₂j " )". Let Z [0;1] be a standard …nite part. Then there exists a …nite partS [0;1]such that every elementz ofZ belongs to S and 8(x₁ , x₂)2S S (jx₁ x₂j " ), i.e. we have B(S; z).

Indeed it su¢ ces to take S = Z. Therefore, the principle of idealization (I) asserts the existence of a …nite part S [0;1] such that any standard element of [0;1]belongs to S and 8(x₁ , x₂) 2S S (jx₁ x₂j " ). Put S =fl₁; l₂; :::; l_ng, wherejl_i+1 l_ij " fori2 f1;2; :::; n 1gand any stan- dard element of [0;1]belongs to S.

Corollary 2.7. For every element l_i of S (S is the set that has been con- structed in the lemma 2.6 ) we have only one of the two cases:

1) l_i is a standard rational number.

2) l_i is outside of " galaxies of all standard rational number.

Proof. Letl_i 2S, then

1) l_i can be a standard rational because S contains all standard elements of [0;1].

2)l_i is not a standard rational thenl_i is not in the" galaxy of any standard rational. Indeed, suppose that l_i = p

q +"$ ($ 6= 0), where p

q is standard.

Then l_i and p

q are elements of S with l_i p

q =j"$j< " which contradicts lemma 2.6 .

Lemma 2.8. For every standard integer n 1. The real numbers x_i of all system fx₁; x₂; :::; x_ng S (S is the set that has been constructed in the lemma 2.6.) are approximated by rational numbers P_i

Q i=1;2;:::;n

to" near with "Q = 0. that is to say:

8<

:

x_i = P_i Q +"

"Q = 0

;i= 1;2; :::; n. (2.6) Proof. Consider the formula:

A(n) "8 fx₁; x₂; :::; x_ng S 9 Pi

Q i=1;2;:::;n

such that:

8<

:

x_i = P_i Q +"

"Q = 0

;i= 1;2; :::; n ".

According to the corollary 2.7, a real x of S is a standard rational or is out- side of " galaxies of standard rationals. In addition, according to lemma 2.5, if x is not in the " galaxy of a rational standard, x is written in the form

8<

:

x = P

Q +"

"Q = 0

. Then in all cases x is written in the form 8<

:

x = P

Q +"

"Q = 0

. Consequently we haveA(1).

Suppose A(n), for a standard integer n, and prove A(n+ 1).

Let (x₁; x₂; :::; x_n; x_n+1) S. Since A is veri…ed for n we have ( xi = p_i

q +" ; i= 1;2; :::; n

"q= 0

. (2.7)

If x_n+1= h₁

k₁ is standard, then becausek₁ is standard and of (2:7) we have 8>

>>

<

>>

>:

x_i = p_ik₁

qk₁ +" = P_i

Q +" ; i= 1;2; :::; n x_n+1 = h₁q

k₁q +":0 = P_n+1 Q +"

"Q="qk₁ = 0

. (2.8)

Let us look at the case where x_n+1 is not a rational standard. In this case the application of the theorem 1.5 to the real qx_n+1 with the in…nitesimal"q implies:

(

qx_n+1 = M

N + ("q)a ("q)N = 0

where a is limited. If a= 0, then from this and (2:7): 8>

>>

<

>>

>:

x_i = p_iN

qN +" = P_i

Q +" ; i= 1;2; :::; n x_n+1 = M

qN +"a = Pn+1

Q +"

"Q ="qN = 0

. (2.9)

Let us look at the case where a is appreciable. Suppose a >0, then 8>

>>

<

>>

>:

x_i = N p_i

N q +" ; i= 1;2; :::; n x_n+1 = M

N q +"a

"N q = 0

. (2.10)

The reduced form of M

N q cannot be a rational standard. Otherwise, x_n+1 and M

N q become two elements ofS such that the separating distance between them, is of the "a form. What, according to lemma 2.6, is not true for two elements of S; for the same reason x_n+1 cannot be in the " galaxy of a standard rational. According to the lemma 2.5:

8<

:

x_n+1= h₁ k1

+" ₁ = h₂ k2

" ₂

"k₁ ="k₂ = 0 ; k₁ =k₂ = +1

(2.11)

Where ₁ 0 and ₂ 0 are two in…nitesimal reals and h₁ k₁, h₂

k₂ are irre- ducibles. Let the element of S succeeding immediately x_n+1 in S (xn+1 <

). Then by lemma 2.6 :

xn+1 ="! = 0,! . The real number x_n+1+

2 is not in the " galaxy of a rational standard, otherwise, x and does not become two successive elements ofS. Hence,

according to the lemma 2.5 ( x_n+1+

2 = s l " ₄

"l = 0 ; l= +1; ₄ 0 and ₄ = 0

. (2.12)

where s

l is irreducible. Let be an unlimited natural number such that p : = 0 and N = E p . Let us take N = max N ; k₂; l . Then N = +1 and is of the E p form with is a positive unlimited real verifying p

: = 0. In the other hand h₂

k₂ and s

l are two elements of F^N such that h₂

k2

, s

l doesn’t contain any rational standard and k₂ = +1 and l = +1. In this situation the lemma 2.4 is applicable and consequently there is a …nite sequence of irreducible rational numbers s_i

li 1 i e

such that h₂

k₂ = s₁ l₁ < s₂

l₂ < ::: < s_e l_e = s

l where e= +1and for i= 1;2; :::; e 1 we have :

s_i+1 li+1

s_i li

=" . Besides we have"l_i = 0,l_i = +1fori= 1;2; :::; e; s_e

l_e s₁

l₁ =" !

2 + _{4 2} . In this paragraph we will associate to eachi 2 f1;2; :::; eg a vector V_i in Qⁿ⁺¹ such that then …rst components ofV_i are in the "-galaxie of then …rst components of (x₁; x₂; :::; x_n; x_n+1), respectively. Whereas the (n+ 1) th component of V_i is equal to s_i

l_i. Indeed, for i = 1 apply lemma 2.2 to the system (l₁x₁; l₁x₂; :::; l₁x_n) with the in…nitesimal "l₁:

8<

:

l₁x_i = T_i;1

t₁ + ("l₁)$ ; i= 1;2; :::; n l₁t₁ = 0

.

Hence 8<

:

x_i = T_i;1 l1t1

+ "$ ; i= 1;2; :::; n

"l₁t₁ = 0

. Then

8>

<

>:

x_i = T_i;1

l₁t₁ + "$ ; i= 1;2; :::; n

x_n+1 = T_n+1;1

l1t1

1

(2.13)

whereTn+1;1 = s₁t₁, "l₁t₁ = 0 and 1 = " ₂. Then we obtain the vector V₁ = T_1;1

l₁t₁ ; T_2;1

l₁t₁ ; :::;T_n+1;1 l₁t₁

T

, where x_n+1 = T_n+1;1 l₁t₁ = s₁

l₁.

Again the application of the lemma 2.2 to the system(l₂x₁; l₂x₂; :::; l₂x_n)with the in…nitesimal8 "l₂, gives:

<

:

l₂x_i = T_i;2

t₂ + ("l₂)$ ; i= 1;2; :::; n l₂t₂ = 0

.

Hence 8<

:

x_i = Ti;2

l₂t₂ + "$ ; i= 1;2; :::; n

"l₂t₂ = 0

. Then 8>

<

>:

x_i = T_i;2

l₂t₂ + "$ ; i= 1;2; :::; n

x_n+1 = T_n+1;2

l₂t₂ ²

(2.14) whereT_n+1;2 =s₂t₂ ,"l₂t₂ = 0and 2 =" ₂+" with0< ₁ < ₂. Then we obtain the vector V₂ = T_1;2

l₂t₂ ; T_2;2

l₂t₂ ; :::;T_n+1;2 l₂t₂

T

, where x_n+1 = T_n+1;2 l₂t₂ = s₂

l₂.

Thus we construct the following vectors:

V_i = T_1;i liti

; T_2;i liti

; :::;T_n+1;i liti

T

; i= 1;2; :::; e (2.15) where for i= 1;2; :::; e: x_n+1 = T_n+1;i

l_it_i ⁱ = s_i

l_i ⁱ with "l_it_i = 0.

Besides 0< " ₂ = ₁ < ₂ < ::: < _e= "!

2 +" ₄ and for i= 1;2; :::; e 1:

i+1 i =" .

Lethbe the smallest integer such thathN q max

i (l_it_i), then"hN q = 0.

On the other hand and according to Robinson’s lemma it exists an integer

W = +1 such that:

"W hN q = 0.

Put K =hN q. From (2:10):

8>

<

>:

x_i = hN p_i

hN q + " = H_i

K + " ; i= 1;2; :::; n

xn+1 = hM

hN q + "a = H_n+1

K + "a

(2.16)

where "K = 0, K max

i (l_it_i).

LetW = min W, !

2 + _{4 2} and T_n+1;i0

l_i0t_i0 be the element of the sequence T_n+1;i

l_it_i i=1;2;:::;e

which is the farthest from T_n+1;1

l₁t₁ verifying T_n+1;i0

l_i0t_i0

T_n+1;1

l₁t₁ ="W

withW W. One notices thatW = +1because by constructionW W = .

Let R 1 be the integer such thatRli0ti0 K <(R+ 1)li0ti0. In this case Rl_i0t_i0 and K are of the same order of magnitude i.e. : K

Rl_i0t_i0 = where is a positive appreciable. Consider, the rationals of the following vector:

RT_1;i0

Rl_i0t_i0, RT_2;i0

Rl_i0t_i0, ..., RT_n;i0

Rl_i0t_i0, RT_n+1;i0 Rl_i0t_i0

T

. (2.17)

Where then …rst components of(x1; x2; :::; xn; xn+1)are in the" galaxies of the n …rst components of the (2.17), respectively. Whereas x_n+1 is far from the last component of (2.17) by "W +" ₂. We will search a positive integer j₀ for which the rational RT_n+1;i0 +j₀H_n+1

Rl_i0t_i0 +j₀K becomes equal to H_n+1

K +"a+"

i.e. equal to xn+1+" . Indeed, put

j = RT_n+1;i0 +jH_n+1 Rl_i0t_i0 +jK

H_n+1

K . (2.18)

Then j =

1 +j where is the distance between RT_n+1;i0

Rl_i0t_i0 and H_n+1 K which is equal to "W +" ₂+"a.

Put 1 +j ="a. For this 1 +j =

"a. Hence

j = 1 "a

"a = 1 "W +" ₂

"a

!

= W + ₂

a = +1

.

Let us takej₀ =E W + ₂ a

!

, hencej₀ = W + ₂

a with 2[0;1[. Then

j0 =

1 +j₀ . After the substitution by the value of and of j0:

j0 = a: "W +" ₂+"a

a+W + ₂ a

= "a W + ₂+a

W + ₂+a a

! .

Hence

j0 = "a:

W + ₂+a

W + ₂+a 1 a

W + ₂+a

!

= "a 1

1 .

Since 1

1 = 1 + , then :

j0 ="a+" . (2.19)

On the other handj₀ and W are of the same order of magnitude; indeed:

j₀ W

= 1

W

W + ₂ a

a

!

= 1 +

a .

Therefore j₀ W

=A with A is appreciable, hence j₀ =AW. Since W W one has: j₀ =AW AW AW.

Lemma 2.9. The denominator of RTi;i0 +j₀Hi

Rl_i0t_i0 +j₀K ( i= 1;2; :::; n+ 1) veri…es

"(Rl_i0t_i0 +j₀K) = 0 and for i= 1;2; :::; n; n+ 1 we have:

x_i = RT_i;i0 +j₀H_i

Rl_i0t_i0 +j₀K +" (2.20)

Proof.

Rl_i0t_i0 +j₀K = K

+j₀K

= K 1

+j₀ . Hence Rl_i0t_i0 +j₀K K 1

+AW . From the fact that"W K = 0; A and are two appreciable numbers, we have "(Rl_i0t_i0 +j₀K) = 0. On the other hand for i = n+ 1 we have from (2.16) x_n+1 = H_n+1

K +"a and from (2.18) and (2.19)

j0 = RT_n+1;i0 +j₀H_n+1 Rl_i0t_i0 +j₀K

H_n+1

K ="a+" . Hence RT_n+1;i0 +j₀H_n+1

Rl_i0t_i0 +j₀K " = H_n+1

K +"a =x_n+1, this means that x_n+1 = RT_n+1;i0 +j₀H_n+1

Rl_i0t_i0 +j₀K +" . For i= 1;2; :::; n we know from(2:17) that:

RT_i;i0

Rl_i0t_i0 x_i ="$. (2.21)

Hence

RT_i;i0 Rl_i0t_i0

H_i

K = RT_i;i0

Rl_i0t_i0 x_i +x_i H_i K RTi;i0

Rl_i0t_i0 x_i + x_i Hi

K ="$+" ="$ .

Therefore

RT_i;i0 Rl_i0t_i0

H_i

K = RT_i;i0K H_iRl_i0t_i0

KRl_i0t_i0 = $. (2.22)

Then we have: RT_i;i0 +j₀H_i Rli0ti0 +j₀K

H_i

K = RT_i;i0K H_iRl_i0t_i0 KRl_i0t_i0 1 +j₀: K

Rl_i0t_i0

= $

1 +j₀ . Since j₀ = +1, then

RT_i;i0 +j₀H_i Rl_i0t_i0 +j₀K

H_i K ="

and seen that for i= 1;2; :::; n, the rational numbers H_i

K are, respectively, in the " halos of x₁; x₂; :::; x_n then:

x_i = RT_i;i0 +j₀H_i Rl_i0t_i0 +j₀K +" . So the lemma is proved.

Since "(Rl_i0t_i0 +j₀K) = 0, then if for i = 1;2; :::; n; n + 1 one takes RTi;i0 +j0Hi

Rl_i0t_i0 +j₀K for Pi

Q then 8<

:

x_i = P_i

Q +" , i= 1;2; :::; n+ 1

"Q= 0

. (2.23)

In the case where a < 0 we take the element of S that precedes x_n+1 i.e.

< x_n+1 (S is ordered) and by doing, to a symmetry near, as we did for the case a >0.

From (2:8), (2:9) and (2:23) we have A(n+ 1). Hence, according to the ex- ternal recurrence principle, the lemma 2.8 is proved.

Let us return to the proof of theorem 2.1 De…ne forZ =fx₁; x₂; :::; x_sg [0;1], the formula:

B(Z) = "9 P_i

Q i=1;2;:::;s

such that :8^stm2N G Z, P_i

Q i=1;2;:::;s

, m

!

"

whereG Z, P_i

Q i=1;2;:::;s

, m

! 8

><

>: 1

" x_i P_i Q

1

m ; = 1;2; :::; s j"Qj 1

m

is in- ternal.

Consider the set

L=fn2N :n jSj &8s 2 f1,:::,ng 8Z =fx₁; x₂; :::; x_sg S :B(Z)g. (2.25) where S is the set that has been constructed in the lemma 2.6 .Then

L=

8>

<

>:

n 2N :n jSj &8s2 f1,:::,ng 8Z =fx₁; x₂; :::; x_sg S, 9 P_i

Q i=1;2;:::;s

8^stm2N G Z, P_i

Q i=1;2;:::;s

, m

! 9>

=

>;. According to lemma 2.8, L (N ) . If L is internal then, according to the Cauchy principle, it must contain (N ) strictly and therefore there is an integer ! = +1 and ! 2 L. If L is external then by the idealization principle (I) we can write L as follows:

L=

8>

<

>:

n 2N :n jSj &8s2 f1,:::,ng 8Z =fx₁; x₂; :::; x_sg S, 8^{stf ini} M 9 P_i

Q i=1;2;:::;s

8m2M G Z, P_i

Q i=1;2;:::;s

,m

! 9>

=

>;. whereM belongs to the set of …nite parts ofN . Therefore, Lis an halo ([4], [6]). Of the fact that(N ) L and no halo is a galaxy (Fehrele principle), then (N )

6

= L. Hence it exists an integer != +1and !2L.

Consequently in the two cases (L internal or external ) we …nds that it exists an integer ! = +1 and ! 2L, this signi…es that! jSj.

By lemma 2.3, there is a …nite part F [0;1] containing all standard elements of [0;1] such that jFj = !⁰ = +1 and !⁰ < !. Then F \S is a …nite part of S containing all standard elements of [0;1] with jF \Sj jFj = !⁰ < !. Put F \ S = fx₁; x₂; :::; x_n0g. Then 9 P_i

Q i=1;2;:::;n0

such

that : 8<

:

x_i P_i Q ="

"Q= 0 ; i= 1;2; :::; n₀

. It follows that if x 2 R is a standard then x E(x) = P_i1

Q +" wherei₁ 2 f1;2; :::; n₀gsince x E(x) is a standard of [0;1]. Hence

x = E(x) + P_i1

Q +" = P_x Q +"

where "Q= 0. Thus the proof is complete.

3 Deduction of the classical equivalent of the main result

The theorem 2.1. can be written as follows

8" 8^str (0< " r) =) 9q 8^stx 8^stt (kqxk< "qt &"q t)

where ", r 2 R ⁺, q 2 N, x 2 R and t 2 R ⁺. By using the idealization principle (I), the last formula is equivalent to

8" 8^str (0< " r) =) 8^{st f ini}X 9q 8(x, t)2X (kqx k< "qt& "q t) where X belongs to the set of …nite parts of R R ⁺. This last formula is equivalent to

8^{st f ini}X8"9^strf(0< " r) =) 9q 8(x, t)2X (kqx k< "qt& "q t)g. Again, by using the idealization principle (I), the last formula is equivalent to

8^{st f ini}X 9^{st f ini} R 8" 9r2R f(0< " r) =) 9q 8(x, t)2X (kqx k< "qt& "q t)g.

where R belongs to the set of …nite parts of R ⁺. By the transfer principle (T), this last formula is equivalent to

8^{f ini}X 9^{f ini}R 8" 9r2R f(0< " r) =) 9q 8(x, t)2X (kqxk< "qt &"q t)g. This last formula is exactly the main theorem announced in the abstract.

Indeed, if X = f(x₁, t₁), (x₂, t₂), ..., (x_n, t_n)g is a …nite part of R R ⁺, then there exist a …nite part R of R ⁺ such that for all " > 0 there exists r 2R such that if0< " r then there exist rational numbers p_i

q i=1;2;:::;n

such that:

8<

:

x_i p_i q "t

"q t

;= 1;2; :::; n.

Ref erences

[1]A. BOUDAOUD,Modélisation de phénomènes discrets et approximations diophantiennes in…nitésimales, Maghreb Mathematical Review Vol 1(1), June 1992.

[2]F.Diener, G.Reeb, Analyse Non Standard, Hermann éditeurs des sciences et des arts . 1989 .

[3] G.H. HARDY & E.M. WRIGHT, An introduction to the theory of num- bers, Clarendon Press, Oxford, 1979.

[4] F. Koudjeti, Elements of External Calculus with an application to math- ematical Finance, PhD thesis, Groninjen (The Netherlands), 1995.

[5]R. Lutz, M.Goze, Non Standard Analysis. A practical guide with applica- tions, Lecture note in Math. N⁰881, Springer Verlag (1981).

[6]E. Nelson, Internal set theory : A new approach to non standard analysis, bull. Amer. Math. Soc. 83(1977) 1165-1198.

[7] W.M. SCHMIDT,Diophantine approximation, Lectures Notes in Mathe- matics, N⁰785, Springer-Verlag, berlin, (1980).

[8] I. P. Van den Berg, Nonstandard Asymptotic Analysis, Volume 1249 of Lecture Notes in Mathematics. Springer Verlag, 1987.